A 58.0 kg skier is moving at 6.00 m/s on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 3.65 m long. The coefficient of kinetic friction between this patch and her skis is 0.310. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 3.50 m high. Required:
a. How fast is the skier moving when she gets to the bottom of the hill?
b. How much internal energy was generated in crossing the rough patch?

a) v = 3.71m/s

b) U = 616.71 J

Explanation:

a) To find the speed of the skier you take into account that, the work done by the friction surface on the skier is equal to the change in the kinetic energy:

(the minus sign is due to the work is against the motion of the skier)

m: mass of the skier = 58.0 kg

v: final speed = ?

vo: initial speed = 6.00 m/s

d: distance traveled by the skier in the rough patch = 3.65 m

Ff: friction force = Mgμ

g: gravitational acceleration = 9.8 m/s^2

μ: friction coefficient = 0.310

You solve the equation (1) for v:

Next, you replace the values of all parameters:

The speed after the skier has crossed the roug path is 3.71m/s

b) The work done by the rough patch is the internal energy generated:

The internal energy generated is 616.71J

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